Question

Consider the population of four juvenile condors. Their weights in pounds are : 2, 3, 6, 10 (a) Let x be the weight of a juvenile condor. Write the possible unique values for x: (NOTE: Separate each value in the list with a comma.) ANS = 2,3,6,10 (b) Find the mean of the population: ANS = 5.25 (c) Let x¯ be the average weight from a sample of two juvenile condors. List all possible outcomes for x¯. (If a value occurs twice, make sure to list it twice.) This is the sampling distribution for samples of size 2: (NOTE: Separate each value in the list with a comma.) ? (d) Find the mean of the sampling distribution: ?

Answer #1

c)

Sample | Average weight |
---|---|

(2,2) | 2 |

(2,3) | 2.5 |

(2,6) | 4 |

(2,10) | 6 |

(3,2) | 2.5 |

(3,3) | 3 |

(3,6) | 4.5 |

(3,10) | 6.5 |

(6,2) | 4 |

(6,3) | 4.5 |

(6,6) | 6 |

(6,10) | 8 |

(10,2) | 6 |

(10,3) | 6.5 |

(10,6) | 8 |

(10,10) | 10 |

The sampling distribution for samples of size 2: 2, 2,5, 2,5, 3, 4, 4, 4.5, 4.5, 6, 6, 6, 6.5, 6.5, 8, 8, 10

d)

Sample mean | Probability |
---|---|

2 | 1/16 = 0.0625 |

2.5 | 2/16 = 0.125 |

3 | 1/16 = 0.0625 |

4 | 2/16 = 0.125 |

4.5 | 2/16 = 0.125 |

6 | 3/16 = 0.1875 |

6.5 | 2/16 = 0.125 |

8 | 2/16 = 0.125 |

10 | 1/16 = 0.0625 |

mean of the sampling distribution = 2 * 1/16 + 2.5 * 2/16 + 3 * 1/16 + 4 * 2/16 + 4.5 * 2/16 + 6 * 3/16 + 6.5 * 2/16 + 8 * 2/16 + 10 * 1/16 = 5.25

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